Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        double[] p = new double[N];
        for (int i = 0; i < N; i++) {
            p[i] = sc.nextDouble();
        }
        double[][] dp = new double[N+1][2*N+1]; //0->-N N->0 2*N -> +N
        dp[0][N] = 1;
        for (int i = 0; i < N; i++) {
            for (int j = 1; j < 2*N; j++) {
                dp[i+1][j+1] += dp[i][j] * p[i];
                dp[i+1][j-1] += dp[i][j] * (1-p[i]);                
            }
        }
        double ans = 0;
        for (int i = N+1; i < 2*N+1; i++) {
            ans += dp[N][i];
        }
        System.out.println(ans);
    }
}